Question

Find the order of the given element.

(c) $\mathbf{\left(}\begin{array}{cc}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}& \frac{\mathbf{1}}{\mathbf{2}}\\ \mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}& \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\end{array}\mathbf{\right)}$in$\mathit{G}\mathit{L}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{ℝ}\mathbf{)}$

Short answer

The order of $\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)$ in $GL(2,ℝ)$ is 3.

Step-by-step solution

Step 1:

The group$\mathit{G}\mathit{L}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{ℝ}\mathbf{)}$is a group of all $\mathbf{2}\mathbf{\times }\mathbf{2}$invertible matrices over the field of real numbers$\mathit{ℝ}$

Order of an element

Order of an element$\mathit{a}\mathbf{\in }\mathit{G}$is defined as a positive integer n such that${\mathit{a}}^{\mathbf{n}}\mathbf{=}\mathit{e}$, where$\mathit{e}$isthe identity element in $\mathit{G}$.

Conclusion

The given element in $GL(2,ℝ)$is $\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)$.Now,

$\begin{array}{l}{\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)}^2=\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)=\left(\begin{array}{cc}-\frac{1}{2}& -\frac{1}{2}\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right),\\ {\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)}^3=\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& -\frac{1}{2}\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\end{array}$

So, the order of $\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)$ in $GL(2,ℝ)$ is 3.